dimension of $L(D)$ of compact riemann surface

Question

Suppose we have some positive divisor on a compact Riemann surface $D$ such that $\text{dim} \ L(D) = 1 + \deg(D)$. I want to show that there exists a point $p \in X$ such that $\dim L(p)=2$ and conclude that we have an isomorphism from $X$ to the Riemann Sphere.

My thoughts so far have been to consider that for any divisor on a Riemann surface $X$, at least one of the following are true. Either $L(D - p) = L(D)$ or that the cxdimension of $L(D-p)$ is 1. I hypothesise that it is the latter, but have been stuck at this part for quite some time.

Answer 1

The following is true. Let $D$ be a effective divisor (so $D \geq 0$). Then we always have $\dim L(D) \leq 1+ \deg D$, with equality iff $D =0$ and $g=0$. The backwards direction is easy, and to do the forwards, I would phrase it in terms of the contrapositive (for some reason I can't get it to work out as stated, but maybe you can), and proceed by induction on the degree of $D$. You'll need some exact sequence relating $L(D)$ and $L(D+p)$.